Using log tables for exponential solutions

Question

I understand how to use a log table to solve something such as $\log(0.00000000453)$ where we would put $(0.000000453)$ into scientific notation, $4.53 \times 10^{-9}$. Then we can use the log table to find the mantissa of the log, which is $0.6561$, and use the characteristic, $-9$ to add together and get $0.6561+(-9)=-8.3439$ and $\log (0.00000000453)=-8.3439$.

However, if I am given $1.64^{28}$, how would I use the log table? I can use log properties and get $28 \times \log 1.64 = 28 \times 0.2148$ (value from log table). But this gives me $6.0144$ which is not $1.64^{28}=1036639.481$. How do I take my log table calculation and get back to the exponential answer?

Answer 1

Since $\log 1.64^{28} = 6.0144$, then $1.64^{28} = 10^{6.0144} = 10^6 \times 10^{0.0144}.$ Find in the tables a certain number $x$ such that $\log x \simeq 0.0144$, and then let $1.64^{28} = x \times 10^6$.

Answer 2

$6.0144$ is the log of $1.64^{28}$.
Then you have to look in the tables in the reverse: as for which number has a log of $0.0144$ (in case interpolating) and then add $\times 10^6$. The all apart of course from truncation errors (the actual $log 1.64^{28}=6.0156277...$)